Melody

I do not see any pics ???

g(x) = 0.5x^2   

is NOT one of those graphs!

The curve labled g is not that function.

I hope you take the time needed to actually learn from my answer  

Find the derivative of the function by using the definition of the derivative.

f(x)=(x+1)/(x-1)

REMEMBER: The first derivative  IS  the gradient of the tangent to the curve

Basically when you find this from first principals you take the gradient of the secant.

I want the gradient of the tangent at (x,f(x))

So I have found the gradient of the SECANT  from  (x,f(x)) to the point   (x+c, f(x+c))

THEN I find the limit of this as c tends to 0

This limit will be  the gradient of the TANGENT at (x,f(x)).

In my case this is

    (That is not written technically at all)

This can be expressed in a couple of different ways.  This is one of them.

\(f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{f(x+c)-f(x)}{(x+c)-x}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{x+c+1}{x+c-1}-\frac{x+1}{x-1}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{(x+c+1)(x-1)}{(x+c-1)(x-1)}-\frac{(x+1)(x+c-1)}{(x-1)(x+c-1)}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{(x+c+1)(x-1)-(x+1)(x+c-1)}{(x+c-1)(x-1)}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{x(x+c+1)-1(x+c+1) -x(x+c-1)-1(x+c-1)}{(x+c-1)(x-1)}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{2x-2(x+c) }{(x+c-1)(x-1)}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{-2c}{(x+c-1)(x-1)}\times \frac{1}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{-2}{(x+c-1)(x-1)}\\ f'(x)=\frac{-2}{(x-1)^2} \)

Check using quotient rule.

f(x)=(x+1)/(x-1)

\(f(x)=\frac{(x+1)}{(x-1)}\\ f'(x)=\frac{(x-1)*1-(x+1)*1}{(x-1)^2}\\ f'(x)=\frac{-2}{(x-1)^2}\\\)

Great :)

Well I am pleased that you answered them yourself Danny.

You should use this forum to help you with questions AFTER you know you need help.

If you have already answered them then you do not need our input.

If you have not seriously attempted them yet then they should not be here or on any other help site.

Maybe you feel overwelmed by maths and that is why you post before you think but it is not a good thing to do.  Not good for you or for us (us = answerers on forums).

You would probably be better off to look for you tube clips where these things are explained.

If you want to learn, rather than have answers freely supplied, there are many good, free teachers who post excellent clips on youtube.

When I want to learn something new, I almost always start with you tube.

I am sure you also now know that it is unacceptable to post a lot of questions all at once.

You should not make a habit of getting answerers here to check your answers either  but if you want to do that occasionally then state that is what you are doing.

Give your logic and your answer.  Don't give the answerers the impression that you have no idea or that you have not even thought about it yet.

I wish you, and everyone on the planet, good luck with your studies. I hope you study hard and acheive well.  

Fin the product of all real solutions to    \(x^{2 - \log_4(x)} = \pi\)

This turned out to be a really cool question!! 

\(x^{2 - \log_4(x)} = \pi\\ log_4[x^{2 - \log_4(x)} ]=log_4[ \pi]\\ (2 - \log_4x) log_4x=log_4 \pi\\ 2log_4x - (\log_4x)^2=log_4 \pi\\ (\log_4x)^2-2(log_4x)+log_4 \pi=0\\ Let A=log_4x\\ A^2-2A+log_4\pi=0\\ A=\frac{2\pm\sqrt{4-4log_4\pi}}{2}\\ A=1\pm\sqrt{1-log_4\pi}\\ sub\; back\;again\\ log_4x=1\pm\sqrt{1-log_4\pi}\\ 4^{log_4x}=4^{1\pm\sqrt{1-log_4\pi}}\\ x=4^{1\pm\sqrt{1-log_4\pi}}\\~\\ \text{So the product of the only two values of x that make this true is}\\ 4^{1+\sqrt{1-log_4\pi}}\times 4^{1-\sqrt{1-log_4\pi}}\\ =4^{1+\sqrt{1-log_4\pi}+1-\sqrt{1-log_4\pi}}\\ =4^2\\ =16 \)

I could not work out anything any more useful either.

Here are the 2 points on the graph

It is good that you finally settled down and allowed yourself to learn something from EP

BUT

Well before you accepted that EP was correct you were certainly not polite to him.

At that point in time there was no thank you from you for the time EP put in, even if an answer is wrong you should still say 'thanks for your time.'

AND

At that point in time you also took points off EP.

I believe you would treat your classroom teacher with considerably more disrespect than you treated EP. 

So I do not think it would be inappropriate for your teacher to have punished you.